Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Terminal Velocity of a Cylinder Freefalling Through a Fluid

  1. May 12, 2016 #1
    Hello :)

    I am an engineer and I am trying to analyse a system which basically contains a cylindrical body free-falling through a body of static water beginning with zero velocity. I am ultimately trying to find what the velocity of the object would be at a depth of 20m. In order to do this I have a spreadsheet set-up which calculates the time derivative velocity, displacement and acceleration terms at any given time.

    The basic equation from which I have derived the time dependant derivatives is as follows:

    Fg-Fd-Fb = m.a (1)

    Where:

    Fg = Bodies 'dry' weight
    Fd = Drag load
    Fb = Buoyancy Force

    The equation can be arrange such that:

    terminal velocity, v(term) = (m/b)*g (2)

    where b is the damping factor.

    It is this damping factor which I am uncertain of. According to Stoke's experiments with a small sphere

    b = 6*pi*mu*r (3)

    where mu =dynamic viscosity

    However this equation does not apply to a cylindrical body is not producing sensible answers.

    I have considered using the equation for drag forces to calculate this as folloiws:

    Fd=0.5*rho*A*Cd*v^2 (4)

    so that, damping factor b becomes:

    b = 0.5*rho*A*Cd*v (5)

    but by using this, the damping factor becomes velocity dependent and therefore varies with time. Therefore the terminal velocity (in accordance with eq. (2)) varies with time which doesn't make sense!

    Has anybody done any similar analysis to this or know what the best approach is to take when solving such a problem? Any help would be greatly appreciated as this has had be stumped for a couple of days now!

    Thanks in advance,

    Ian :)
     
  2. jcsd
  3. May 12, 2016 #2

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    There is no real reason to bring some magical damping coefficient into play, just a magical drag coefficient. Start with the force balance and think about what terminal velocity actually means in terms of the forces. Since you already know how to write expressions for each force, you ought to be able to come up with an answer fairly quickly.
     
  4. May 12, 2016 #3

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    This problem is extremely complex, because the cylinder can tumble or otherwise execute complex motion. I had a hard time finding a good reference, but these may be of use to you:
    http://www.sciencedirect.com/science/article/pii/S0889974615002662
    http://faculty.nps.edu/pcchu/web_paper/proceedings/wit/afm2002_chu.pdf
    http://www.fast.u-psud.fr/~hulin/papers/POF_cylinder_Dangelo.pdf
     
  5. May 13, 2016 #4
    Thanks very much Andy. I'll have a read through these this morning. Wish me luck.

    I probably should have also mentioned that I am assuming only one degree of freedom, so that the cylinder can only move in the vertical direction and cannot translate or rotate. Although this assumption doesn't really apply to a simple cylinder, what I am trying to model is offshore structure which will be weighted in such a way that rotation will be not permitted.

    Cheers again.
     
  6. May 13, 2016 #5
    Hi. Thanks for the response.

    The approach I'm taking is the modelling it as a classical Newtonian system so that it is represented by a mass-spring-damper as follows:

    F(t) = m.a = M.g + b.v + k.x (1)

    where the stiffness term is assumed to be zero as there is no restoring force. At terminal velocity, the above equation becomes equal to zero because the acceleration is zero at terminal velocity so that:

    0 = M.g + b.v(term) (2)

    Now by rearranging this I can find the terminal velocity but I believe I do require the damping factor, b, in order to do so. Note the damping factor is not a coefficient and has units of kg/s. Hopefully that makes a bit more sense!

    Thanks again :)
     
  7. May 13, 2016 #6

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    If I understand you, the flow is perpendicular to the cylinder axis, and you are effectively treating this as a 2-D problem (fluid flow past an infinite cylinder)? Is this Stokes flow (low Reynolds number), or are you concerned with arbitrary Reynolds numbers?

    The case of laminar flow around an infinite cylinder is a 'standard problem' and while more difficult than flow around a sphere, there is an analytic expression for the fluid drag per unit length:

    F = (4πU)/(.5-γ-ln(2aU/8ν)), where U is the (distant) fluid velocity, γ is Euler's constant (0.577...), a the cylinder radius, and ν the kinematic viscosity of the fluid. This expression was originally derived by Oseen and is in Lamb's "Hydrodynamics". There, the full solution is derived and also generalized to flow past an ellipse. Even though you are treating this as an infinite cylinder, the endcaps can have a significant portion of the total fluid drag, and it matters if the ends are squared-off or hemispherical or some other shape.

    For higher Reynolds numbers when there is flow separation, you are dealing with von Karman vortex streets and that's beyond my experience.

    If, on the other hand, you are thinking about flow parallel to the cylinder axis, That is also a solved problem- it's essentially the same as flow past a flat plate and the plate is 2πa across.

    Does this help?
     
  8. May 13, 2016 #7

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    Ah, I see what you were doing now. It's unorthodox but I suppose it makes some sense. Personally, I think it is easier to just do a simple force balance:
    [tex]\Sigma F = ma = F_g - F_b - F_d,[/tex]
    where ##F_g## is the force of gravity, ##F_b## is the force of buoyancy, ##F_d## is the force of drag, and the positive direction is down. You obviously know that ##F_g = mg##. I would argue that it is best (since you are including buoyancy) to case everything in terms of diameter and density, so
    [tex]F_g = \rho_{c}V_{c}g = \frac{\pi}{4} d^2 L \rho_{c} g,[/tex]
    where ##L## is however long your cylinder is. Buoyancy is also known, and is the weight of the displaced fluid,
    [tex]F_b = \frac{\pi}{4} d^2 L \rho_{f} g.[/tex]

    Drag is fairly easy to estimate for a cylinder. First, though, you have to guess whether or not you are in the Stokes flow regime, or the more classical regime where drag is proportional to the square of velocity. You are almost certainly not in the Stokes regime, so you are probably safe in assuming that drag is proportional to ##v^2##, but you can check this after you solve it by calculating the Reynolds number, ##Re_d = \frac{\rho_f v d}{\mu}##, and if ##Re_d \ll 1##, then you will have to switch it up and use a Stokes drag model. I doubt this applies, though. Instead, just use
    [tex]F_d = \dfrac{1}{2} \rho_f v^2 C_d A.[/tex]
    For a cylinder, your relevant ##A## is just the frontal area, so ##A = Ld##. You can find reasonable values for the ##C_d## of a cylinder all over the place.

    So that fills in all of your missing terms, and as you've already recognized, terminal velocity implies that ##a = 0##, so you have velocity occutring only one time in the three remaining terms. It should be fairly simple to simply rearrange the equation to solve for ##v##. That ##v## is your terminal velocity. If you are wanting to know how fast it reaches that speed, then you would have to solve it as a differential equation without zeroing out the ##a## term.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Terminal Velocity of a Cylinder Freefalling Through a Fluid
  1. Terminal Velocity ? (Replies: 5)

  2. Terminal velocity (Replies: 9)

Loading...